Let A = 1-6, B = 1-6, C = 1-6, and n (x ~ y) means that a, b and c all belong to {x, x+ 1, .., y}.

Then n (e = 0) = 6 * [n (1)] = 6 *1= 6,

p(e=0)=6/2 16

N(e = 1)= 5 *[N( 1 ~ 2))-N( 1)-N(2)]= 5 *(8-2)= 30，

p(e= 1)=30/2 16

N(e = 2)= 4 *[N( 1 ~ 3)-N( 1 ~ 2)-N(2 ~ 3)+N(2)]= 4 *(27-8-8+ 1)= 48，

p(e=2)=48/2 16

N(e = 3)= 3 *[N( 1 ~ 4)-N( 1 ~ 3)-N(2 ~ 4)+N(2 ~ 3)]= 3 *(64-27-27+8)= 54，

p(e=3)=54/2 16

N(e = 4)= 2 *[N( 1 ~ 5)-N( 1 ~ 4)-N(2 ~ 5)+N(2 ~ 4)]= 2 *( 125-64-64+27)= 48，

p(e=4)=48/2 16

N(e = 5)= 1 *[N( 1 ~ 6)-N( 1 ~ 5)-N(2 ~ 6)+N(2 ~ 5)]= 2 16- 125- 125+64 = 30，

p(e=5)=30/2 16

To sum up, E(e)=2.9 1667.